0AFE:02B0 00 00 00 00 00 2C 44 63-71 1F 71 1F 00 00 8001 .....,Dcq.q..... 0AFE:0

Question

0AFE:02B0 00 00 00 00 00 2C 44 63-71 1F 71 1F 00 00 8001   .....,Dcq.q.....

0AFE:02C0 01 00 07 FE FF FF 3F 00-00 00 EC ED E1 04 00FE   ......?.........

0AFE:02D0 FF FF 0F FE FF FF 2B EE-E1 04 96 9C BF 0D 0000   ......+.........

0AFE:02E0 00 00 00 00 00 00 00 00-00 00 00 00 00 00 0000   ................

0AFE:02F0 00 00 00 00 00 00 00 00-00 00 00 00 00 00 55AA   ..............U.

Using the above memory dump of the data part of a partitiontable, find the following information for 1st and2nd partitions.

Explanation / Answer

->Each partition table entry is 16 bytes and containsthings like the start and end location of a partition.T
->The start size in sectors, the partition "type" and the"active"flag. ->There is no central clearing house to assign the codesused in the one byte "type" field. But codes are assigned (orused) to define most every type of file system that anyone has everimplemented on the x86 PC: 12-bit FAT, 16-bit FAT, HPFS, NTFS,etc. Plus, an extended partition also has a unique typecode. ->The 16 bytes of a partition table entry are used asfollows:     +--- Bit 7 isthe active partition flag, bits 6-0 are zero.   |

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